Solving Radical Inequalities Part 2 Lesson

In our last lesson, we learned a basic strategy that can be used to solve a radical inequality. In this lesson, we will use the same method and work with a more challenging problem type, one where variables occur outside of the radical. As we mentioned previously, the method that is described below is not necessarily the fastest, especially if you have an easy problem. The benefit of this method is that it will work for any scenario. We previously discussed a problem that can arise when squaring both sides of an inequality. $$-9 < 8 \hspace{.15em}\text{true}$$ Let's square both sides: $$81 < 64 \hspace{.15em}\text{false}$$ We can see that we must be extra careful when squaring both sides of an inequality. To avoid these types of issues, we will use a different approach and work with the related equation.

Solving a Radical Inequality:

• Find any domain restrictions
  • Unless we are told otherwise, we will assume we are working with real numbers
  • We can't have a negative radicand with an even index
• Solve the related equation
  • Replace the inequality symbol with an equality symbol
• Split the number line up into intervals based on the critical values
  • Domain restriction
  • Solution(s) of the related equation
• Test numbers in each interval to obtain your solution set
• Consider the critical values separately

Example 1: Solve each inequality. $$\sqrt{x - 3}< 2x - 12$$ Step 1) Find any domain restrictions:
Set your radicand (x - 3) as greater than or equal to zero and solve the inequality: $$x - 3 ≥ 0$$ $$x ≥ 3$$ Step 2) Replace the inequality symbol with an equality symbol and solve the resulting equation: $$\sqrt{x - 3}=2x - 12$$ $$x=7$$ This solution gives us a boundary. We can use it to split the number line up into three intervals. A region that is less than 3, where no values will work due to the domain restriction. A region that is between 3 and 7, and a region that is greater than 7. For the numbers 3 and 7, we will consider them separately. Step 3) Test numbers in each interval to obtain your solution set:
We know that nothing in interval A (values less than 3) works due to the domain restriction. For interval B, we can test 4: $$\sqrt{4 - 3}< 2(4) - 12$$ $$1 < -4$$ This statement is false, so we know that numbers in interval B or values between 3 and 7 won't work as a solution. For interval C, we can test 12. $$\sqrt{12 - 3}< 2(12) - 12$$ $$3 < 12$$ This statement is true, so we know that numbers that are greater than 7 will work as a solution. Now, let's consider our critical values of 7 and 3. Anything less than 7 does not work and so 3 will not be part of the solution set. Additionally, when we consider 7, it is excluded here since we have a strict inequality and 7 is the solution for the related equation. We will report our answer as: $$x > 7$$ $$(7, \infty)$$ Example 2: Solve each inequality. $$\sqrt{25x^2 - 10x + 1}< 3x + 4$$ Step 1) Find any domain restrictions:
Set your radicand (25x² - 10x + 1) as greater than or equal to zero and solve the inequality:
Note: $$25x^2 - 10x + 1=(5x - 1)^2$$ $$(5x - 1)^2 ≥ 0$$ As we previously discussed in our lesson on solving quadratic inequalities, we really don't have to do much work here since the left side is squared. We know that squaring a number produces a non-negative result. This means (5x - 1)² will produce either 0 or some positive value. Therefore, our domain is the set of real numbers. $$x ∈ ℝ$$ Step 2) Replace the inequality symbol with an equality symbol and solve the resulting equation: $$\sqrt{25x^2 - 10x + 1}=3x + 4$$ $$x=-\frac{3}{8}, \frac{5}{2}$$ These solutions give us our boundaries. We can use them to split the number line up into three intervals. A region that is less than -3/8, a region that is between -3/8 and 5/2, and a region that is greater than 5/2. For the numbers -3/8 and 5/2, we will consider them separately. Step 3) Test numbers in each interval to obtain your solution set:
For interval A, we can test -1: $$\sqrt{(5(-1) - 1)^2}< 3(-1) + 4$$ $$6 < 1$$ This statement is false, so we know that numbers in interval A or values less than -3/8 won't work as a solution. For interval B, we can test 1: $$\sqrt{(5(1) - 1)^2}< 3(1) + 4$$ $$4 < 7$$ This statement is true, so we know that numbers in interval B or values between -3/8 and 5/2 will work as a solution. For interval C, we can test 3: $$\sqrt{(5(3) - 1)^2}< 3(3) + 4$$ $$14 < 13$$ This statement is false, so we know that numbers in interval C or values greater than 5/2 won't work as a solution. Now, let's consider our critical values of -3/8 and 5/2. We know these values solve the related equation. Here, we have a strict inequality so both will be excluded from the solution. We will report our answer as: $$-\frac{3}{8}< x < \frac{5}{2}$$ $$\left(-\frac{3}{8}, \frac{5}{2}\right)$$ Example 3: Solve each inequality. $$\sqrt[3]{7x + 6}≥ -2x + 9$$ Step 1) Find any domain restrictions:
In this case, we have a cube root, which means our index is odd. In the real number system, we can take the cube root of a negative number, therefore, our domain is the set of real numbers. $$x ∈ ℝ$$ Step 2) Replace the inequality symbol with an equality symbol and solve the resulting equation:
$$\sqrt[3]{7x + 6}=-2x + 9$$ Note: If you want to solve this equation on your own, you can cube both sides: $$7x + 6=(-2x + 9)^3$$ $$8x^3 - 108x^2 + 493x - 723=0$$ Note: We haven't yet covered how to factor this type of polynomial yet. We need to learn about the rational roots test, which will be covered much later in the course. For now, we can take the factored form as given. $$(x - 3)(8x^2 - 84x + 241)=0$$ $$x=3$$ Note: The quadratic above will produce two non-real complex solutions, which we will not use here since we are working with real numbers only.
The solution gives us our boundary. We can use it to split the number line up into two intervals. A region that is less than 3 and a region that is greater than 3. We will consider 3 separately. Step 3) Test numbers in each interval to obtain your solution set:
For interval A, we can test 0: $$\sqrt[3]{7(0) + 6}≥ -2(0) + 9$$ $$\sqrt[3]{6}≥ 9$$ Note: $$\sqrt[3]{6}≈ 1.82$$ This statement is false, so we know that numbers in interval A or values less than 3 won't work as a solution. For interval B, we can test 4: $$\sqrt[3]{7(4) + 6}≥ -2(4) + 9$$ $$\sqrt[3]{34}≥ 1$$ Note: $$\sqrt[3]{34}≈ 3.24$$ This statement is true, so we know that numbers in interval B or values greater than 3 will work as a solution. Now, let's consider our critical value of 3. We know this value solves the related equation. Here, we have a non-strict inequality so it will be included in the solution. We will report our answer as: $$x ≥ 3$$ $$[3, ∞)$$

---

Skills Check: